什么是二项式概率分布示例？

概率是某事发生的可能性。当我们说某事的概率时，它意味着某事的可能性有多大。有些事件的概率很高，很可能发生，有些事件的概率较小，这意味着它们不太可能发生。示例 – 投掷无偏硬币得到尾巴的概率是 1/2，掷骰子得到大于 4 的数字的概率是 1/3。换句话说，它是事件作为实验结果发生的机会的量度。

概率论是一种非常强大的工具，用于组织、解释和应用信息，在数据科学、交易、马匹投注等各个领域都非常有用。

二项式实验

二项式实验是一个实验，其中有固定数量的试验（例如 n），每个试验都独立于其他试验，只有 2 个结果：成功或失败，并且每个结果的概率在试验之间保持不变。二项式实验的例子，

Binomial Experiment	Possible Outcomes
Did 2 appear on the dice?	Yes / No
Condition of a statement	True / False
Tossing a coin	Heads / Tails

二项式概率

当给定任何一个二项式实验，其中我们进行多次随机实验（例如，掷硬币 7 次或掷骰子 10 次），那么在 n 次试验中找出某个结果的概率称为它的二项式概率.假设一个二元实验有n 个独立的试验，有两个结果：

• 成功
• 失败

现在在n次试验中获得r次成功的概率是：

P = ^nC_r.p^r.q^n-r

其中p = 成功概率， q =失败概率，使得p + q = 1。

对称二项分布的图形表示

二项分布由多个伯努利事件组成。伯努利分布和二项分布的区别在于，伯努利分布的期望值给出了单次试验的预期结果，而二项分布的期望值表明了预期获得特定结果的次数。二项分布图表示达到我们期望结果特定次数的可能性。

示例问题

问题 1：如果一个无偏的硬币被抛 7 次，那么找出恰好得到 3 个正面的概率。

解决方案：

Given number of trials(n) = 7, number of success(r)= 3

= Probability of success = Probability of getting a head in a trial (p) = 1/2

= Probability of failure = Probability of not getting a head in a trial (q) = 1/2

Now applying binomial probability formula:

= Probability of getting exactly 3 heads (P) = ⁿC_r.p^r.q^{n – r}

= ⁷C₃.(1/2)³ × (1/2)⁴

= (7!/3!.4!) × (1/2)⁷

= 35/128 

问题2：一个骰子被掷了5次，得到一个奇质数被认为是成功的。找出概率：

1. 正是一次成功
2. 至少一次成功

解决方案 ：

Given:  Number of trials (n) = 5,

Number of odd prime numbers from 1 to 6 = 2 (3, 5)

= Probability of success = Probability of getting a 3 or 5 on the dice(p) = 2/6 = 1/3

= Probability of failure = Probability of not getting 1, 2, 4, 6 on the dice(q) = 1 – 1/3 = 2/3

• Exactly one success(r = 1)

 Applying binomial probability formula:

=> Probability of getting exactly 1 success (P) = ⁿC_r.p^r.q^n-r

= ⁵C₁. (1/3)¹. (2/3)⁴

= 80/243 (Ans)

• At least one success( r = 1, 2, 3, 4, 5)

Now since it is given at least one succes, add all the binomial probabilities for r = 1, 2, 3, 4, 5

Applying binomial probability formula:

= Probability of getting at least 1 success (P) = P(r = 1) + P(r = 2) + P(r = 3) + P(r = 4) + P(r = 5)

(getting 1 success) (2 success) (3 success) (4 success) (5 success)  

= 1 – P(r = 0) (getting no success)    

= 1 – ⁿC₀.p⁰.q^{n – 0}                                         

= 1 – ⁵C₀.(1/3)⁰.(2/3)⁵

= 1 – 32/243                               

= 211/243 

问题 3：小丑从一副洗好的 52 张牌中抽出 4 张牌并替换。求他从套牌中至少抓到 3 张王的概率。

解决方案：

Given: Number of cards to be drawn(n) = 4

Success =  getting a king 

Probability of getting a king card from 52 random cards(p) = 4/52 = 1/13 (Since total no of kings = 4 and each card is replaced after every pick)

Probability of failure(q) = 1 – 1/13 = 12/13 

= Probability of getting at least 3 king in this case = P(r = 3) + P(r = 4)

(getting 3 kings) (getting 4 kings)

Applying binomial probability formula = ⁴C_3.(1/13)³.(12/13)¹ + ⁴C₄.(1/13)⁴.(12/13)⁰

= (1/13)³.(4.12/13 + 1/13) (Taking common on both sides)

= 49/28561 

问题 4：您进行的 MCQ 测试只有 5 个问题。每道题有4个选项，每道题可以有一个或多个正确答案。由于您没有为测试学习任何内容，因此您决定随机标记所有答案。找出您在 5 道题中恰好答对 3 道题的概率，才能通过考试。

解决方案：

Given: Number of questions(n) = 5

Now to find the probability of success, first, find the total 

no of the ways a question can be answered.

As shown in the figure above, there are 4 cases :

= number of ways to answer a question when only 1 option is correct = ⁴C₁ = 4 ways 

= number of ways to answer a question when 2 options are correct = ⁴C₂ = 6 ways 

= number of ways to answer a question when 3 options are correct = ⁴C₃ = 4 ways 

= number of ways to answer a question when 4 options are correct = ⁴C₄ = 1 way 

Adding up all ways, the total no of ways = 15 ways.

Now out of these 15 ways, only one will be correct for a particular question.

Thus, the probability of success(i.e. getting a correct answer) = 1/15

And the probability of failure = 1 – 1/15 = 14/15

Now,

The probability that you get exactly 3 question correct out of 5

= ⁵C₃.(1/15)³.(14/15)

= 10. (196/759375)              

= 392/151875